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<a name="Copy-Assignment"></a>
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<a name="Implicit-Copy_002dAssignment-for-Virtual-Bases"></a>
<h4 class="subsection">14.7.4 Implicit Copy-Assignment for Virtual Bases</h4>

<p>When a base class is virtual, only one subobject of the base class
belongs to each full object.  Also, the constructors and destructors are
invoked only once, and called from the most-derived class.  However, such
objects behave unspecified when being assigned.  For example:
</p>
<div class="smallexample">
<pre class="smallexample">struct Base{
  char *name;
  Base(char *n) : name(strdup(n)){}
  Base&amp; operator= (const Base&amp; other){
   free (name);
   name = strdup (other.name);
  }
};

struct A:virtual Base{
  int val;
  A():Base(&quot;A&quot;){}
};

struct B:virtual Base{
  int bval;
  B():Base(&quot;B&quot;){}
};

struct Derived:public A, public B{
  Derived():Base(&quot;Derived&quot;){}
};

void func(Derived &amp;d1, Derived &amp;d2)
{
  d1 = d2;
}
</pre></div>

<p>The C++ standard specifies that &lsquo;<samp>Base::Base</samp>&rsquo; is only called once
when constructing or copy-constructing a Derived object.  It is
unspecified whether &lsquo;<samp>Base::operator=</samp>&rsquo; is called more than once when
the implicit copy-assignment for Derived objects is invoked (as it is
inside &lsquo;<samp>func</samp>&rsquo; in the example).
</p>
<p>G++ implements the &ldquo;intuitive&rdquo; algorithm for copy-assignment: assign all
direct bases, then assign all members.  In that algorithm, the virtual
base subobject can be encountered more than once.  In the example, copying
proceeds in the following order: &lsquo;<samp>val</samp>&rsquo;, &lsquo;<samp>name</samp>&rsquo; (via
<code>strdup</code>), &lsquo;<samp>bval</samp>&rsquo;, and &lsquo;<samp>name</samp>&rsquo; again.
</p>
<p>If application code relies on copy-assignment, a user-defined
copy-assignment operator removes any uncertainties.  With such an
operator, the application can define whether and how the virtual base
subobject is assigned.
</p>



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